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Factorisation is defined as dividing an integer or polynomial into factors which when multiplied together, result in the initial integer or polynomial. We use the factorisation method to simplify any algebraic or quadratic equation by representing it as the product of factors rather than expanding the brackets. Any equation's factors can be an integer, a variable, or the algebraic expression itself.
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Key Takeaways: Factoring, Factorize, Factorization, Polynomials, Matrix, Factors
Also read: Isosceles Triangle Theorems
What is Factorisation?
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Factorisation is the process of converting a number or a polynomial into a product of many factors of other polynomials, which when multiplied yields the original number.
Use the factorisation formula to factorize a number. The process of converting one entity (for example, a number, a matrix, or a polynomial) into a product of another entity, or factors, which when multiplied together yields the original number.
Also read: Definite Integral Formula
What is a Factor?
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Factors are numbers, algebraic variables, or algebraic expressions that divide the number or expression without leaving any remainder.
For example, the factor of 9 is 1,3,9.
Factorisation in Mathematics
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Factorisation is the process of writing an algebraic expression as a product of its factors. These variables, numbers, or algebraic expressions can be used as factors.
A number, to the factor, means dividing it into numbers that can be multiplied to get the original number. For example,
| 24 = 4 × 6 | 4 and 6 are the factors of 24 |
| 9 = 3 × 3 | 3 is the factor of 9 |
Numbers can also be factored into various combinations. There are several methods for determining the Factors of a Number. Finding the factors of an integer is simple, but finding the factors of algebraic equations is more difficult. So, let's figure out how to find the factors of a quadratic polynomial.
Also read: Differential Equation
Factorisation in Algebra
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Since 12 is divided without a remainder, the numbers 1, 2, 6, and 12 are all factors of 12. It is an important algebraic process that is used to simplify expressions, fractions, and solve equations. Algebra factorization is another name for it.
Factors and Terms
What exactly is a term?
It is something that will be added or subtracted in an expression (subtracting is adding a negative number).
If 2x + 7 is an expression, then the terms are 2x and 7.
Sum= term + term
What exactly is a Factor?
It is a number that will be multiplied in an expression.
factor factor factor factor factor factor factor factor factor factor factor factor factor factor factor factor factor factor factor
As an example:
4(2q – 6) = p
The factors are 4 and 2q – 6, while the variables are 2q and 6.
Basic Factorisation Formula
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a2– b2 = (a – b)(a + b)
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
a3 – b3 = (a – b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 – ab + b2)
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca
Also Read: Division of Polynomial by Another Monomial
Factorisation in Methods
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There are four methods to factorise the algebraic expressions.
- Common factors method
- Regrouping terms method
- Factorisation using identities
- Factors of the form (x+a) (x+b)
Common Factors Method
We simply extract the common factors among each term of the given expression using this method.
3x + 9 is an example of a factorisation.
Because 3 is the common factor for both the terms 3x and 9, we get 3x + 9 = 3(x+3) when we take 3 as a common factor.
Regrouping of Terms Method
Regrouping means rearranging the given expression based in the like terms or similar terms.
For example, 2xy + 3x + 2y + 3 can be rearranged as:
2xy + 3x + 2y + 3
Expanding the terms into factor form.
= 2 × x × y + 3 × x + 2 × y + 3
Re-arrange to get the common factor
= x × (2y + 3) + 1 × (2y + 3)
Now (2y + 3) is the common ffactor we can take out.
= (2y + 3) (x + 1)
Thus, these are the required factor s.
Factorisation using Identities
By using the common identities, we can factirise the given expression.
Example: Factorise 4x2 – 9.
Solution: By using the algebraic identities, we know;
a2 – b2 = (a – b) (a + b)
Hence, we can write,
4x2 – 9
= (2x)2 – 32
= (2x + 3) (2x – 3)
Note: For the rest of the identities check with the formulas for factorisation mentioned in the above section.
Also read: Difference between Sequence and Series
Factors of the Form (x+a) (x+b)
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If a given expression is in the form of x2 + (a + b) x + ab, then the factors will be (x+a) and (x + b).
Example: Factorise x2 + 5x + 6.
Solution: If we compare the given expression with x2 + (a + b) x + ab, then;
a+b = 5 ….(i)
ab = 6 ….(ii)
Now we will try putting the values for a and b such that they satisfy the above two equations.
If we put a = 4 and b = 1, then a + b = 5 satisfies the eq.(i) but a.b = 4, does not satisfy eq.(ii).
again, if ? = 2 and b = 3, then a + b = 5 satisfies eq.(i) but a.b = 6, does not satisfy eq.(ii).
x2 + 5x + 6
= x2 + (2 + 3)x + 2.3
= x2 + 2x + 3x + 2.3
= x(x+2) + 3(x+2)
= (x+2) (x+3)
Also Read: Polynomial Important Questions
Factorization Formula For a Quadratic Polynomial
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A "quadratic" polynomial is one that looks like "ax² + bx + c," where "a," "b," and "c" are just numbers. For a simple case of factoring, identify the two numbers that will not only multiply to equal the constant term "c," but will also add up to equal "b," the x-term coefficient.
When solving quadratic polynomials, factoring formulas algebra is especially important. When reducing formulas, we usually have to remove all of the brackets, but in some cases, such as with fractional formulas, we can use factorisation to shorten the formula.
Things to Remember
- Factorisation is the process of writing an expression as a product of its factors.
- A prime factor is an irreducible factor that cannot be expressed further as a factor product.
- These identities make it simple to factorize some expressions:
a2 + 2ab + b2 = (a + b)2
a2 - 2ab + b2 = (a - b)2
a2 - b2 = (a - b)(a + b)
x2 + (a + b)x + ab = (x + a)(x + b)
- The number 1 is a factor of every algebraic term, but it is only shown when it is required.
- When factoring x2 + (a + b)x+ ab by splitting the middle term, the two numbers that give the product ab and (a + b) as the coefficient of x must be carefully chosen with the correct sign.
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Sample Questions
Ques. Factorise: 9x + 18y + 6xy + 27 [3 Marks]
Ans: Here, we have a common factor 3 in all the terms.
9x + 18y +6xy + 27 = 3[3x + 6y + 2xy + 9]
We find that 6y = 2xy + 9 = 1(2xy + 9)
i.e. a common factor in both the groups does not exist,
Thus, 3x + 6y + 2xy + 9 cannot be factorised.
In regrouping the terms, we have
6y + 2xy + 9 = 3x + 9 + 2xy + 6y
= 3(x + 3) + 2y(x + 3)
= (x + 3)(3 + 2y)
Now, 3[3x + 6y + 2xy + 9] = 3[(x + 3)(3 + 2y)]
Thus, 9x + 18y + 6xy + 27 = 3(x + 3)(2y + 3)
Ques: Factorise the Quadratic polynomial: x2 + 7x + 6. [2 Marks]
Ans: The constant term is 6, which can be written as the product of 2 and 3 or of 1 and 6.
But 2 + 3 = 5, hence, 2 and 3 are not the numbers needed in this case,
On the other hand 1 + 6 = 7, hence 1 and 6 can be used:
X² + 7x + 6 = (x + 1)(x + 6)
Note that the order dpesn’t matter in multiplication, so the above answer can be written as (x + 6)(x + 1).
Ques: Factorise: x2 – 64. [2 Marks]
Ans: Given, x2 – 64
We can also write the given expressipn as:
⇒ x2 – 82 [Since 8 x 8 = 64]
Now by using the formula,
⇒ a2 – b2 = (a + b) (a – b)
⇒ x2 – 82 = (x + 8) (x – 8)
Ques: Find the greatest common factor of 6x7 + 3x4 − 9x3. [2 Marks]
Ans: Solution: Given, 6x7 + 3x4 − 9x³
Now factoring the given expression, we get;
⇒ 3x3 (2x4 + x − 3)
Therefore, the greatest common factor is 3x3.
Ques: Factorise (7x + 7x3) + (x4 + x6). [2 Marks]
Ans: (7x + 7x3) + (x4 + x6)
Open all the brackets:
⇒ 7x + 7x3 + x4 + x6
Taking the common factor,
⇒ 7x(1 + x2) + x4(1 + x2)
⇒ (7x x4)(1 + x2)
⇒ x(7 + x3)(1 + x2)
Hence, the required factors.
Ques: Factorise the following polynomials. [2 Marks]
(a) 6p(p – 3) + 1 (p – 3)
(b) 14(3y – 5z)3 + 7(3y – 5z)2
Ans: (a) 6p(p – 3) + 1(p – 3) = (p – 3)(6p + 1)
(b) 14(3y – 5z)3 + 7(3y – 5z)2
= 7(3y – 5z)2 [2(3y – 5z) +1]
= 7(3y – 5z)2 (6y – 10z + 1)
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